# Energy and Mass Analysis of a Control Volume

## Homework Statement

Consider a system in which R 134-A is moved through an evaporator. The evaporator removes energy from the the surroundings at a rate of 0.54 kW. The pressure = 120 kPa and x = 0.2 right before the fluid enters the evaporator. After exiting the evaporator, the pressure = 120 kPa and T = -20 C. What is the mass flow rate through the evaporator?

## Homework Equations

(1) $$\sum$$mass_in = $$\sum$$mass_out

(2) $$\dot{q}$$ - $$\dot{w}$$ = $$\sum$$(($$\dot{m}$$_out) * (h + [(V_2)^2/2] + gz)) - $$\sum$$(($$\dot{m}$$_in) * (h + [(V_1)^2/2] + gz))

(3) $$\dot{E}$$ = $$\dot{m}$$e
where e = u + ke + pe​

## The Attempt at a Solution

Assuming steady flow, equation 1 is applicable, so the mass flow rate going into the evaporator is the same as the mass flow rate exiting the evaporator. However, I'm not exactly sure how to go about from there.

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